Let $\mathcal L$, $\mathcal G$ : $V$ $\rightarrow$ $V$ be two linear operators, prove the following Let $\mathcal L, \mathcal G : V \rightarrow V$ be two linear operators, so that $\mathcal L^2 = \mathcal G^2$ and $\text{Ker }$$\mathcal L \cap\text{Ker } \mathcal G=\{0\}$. Prove:
1) $\mathcal L (\text{Ker }\mathcal G) \subset \text{Ker }\mathcal L$
2) $\dim\mathcal L(\text{Ker }\mathcal G) = \dim\text{Ker }\mathcal G$
3) $\text{rank }\mathcal L=\text{rank }\mathcal G$
4) If $\mathcal L \circ\mathcal G = \mathcal G\circ\mathcal L$ then $\text{Ker }\mathcal G \subset \text{Ker }(\mathcal G\circ\mathcal L)$
What have I considered so far:
1) $\mathcal L(\text{Ker }\mathcal G) = \mathcal L (\text{Im}\{\text{Ker } \mathcal G\})=\mathcal L (\{0\}) =  \overrightarrow 0 \in \text{Ker }\mathcal L$
2) Let $\dim \text{Ker }\mathcal G = r$, since $\text{Ker }(\mathcal L)\cap \text{Ker }(\mathcal G)= \overrightarrow 0$, that means that $\mathcal L$ will map all vectors from set of vectors that make $\text{Ker }\mathcal G$ onto some image of those vectors, which has also dimension of $r$. Is this way of thinking correct?
3) I am not sure how to prove this one. I can maybe say that is true because rank is image of those vectors that are not in $\text{Ker}$, and draw it, but I am not sure that would be enough.
4)   Since $\mathcal G$ would map $\overrightarrow 0$ vector back to $\text{Ker } \mathcal G$ then $\text{Ker }(\mathcal G\circ\mathcal L)$ is  $\text{Ker }\mathcal G$? But here is says it has to be subset.
Is this way of thinking correct? I think 3) and 4) are for sure not.
Thank you.
 A: 1) Let $x \in \mathcal L(\ker \mathcal G)$, so $x=\mathcal L( y), y \in \ker \mathcal G $, thus we have $\mathcal L (x)=\mathcal L^2(y)=\mathcal G^2(y)=\mathcal G (0)=0$. So $x \in \ker \mathcal L$, so $\mathcal L(\ker \mathcal G) \subset \ker \mathcal L $.
2) A neat way to prove it is to use Rank–nullity theorem. Let $u$ be the restriction of $\mathcal L$ to $\ker \mathcal G$. By the Rank–nullity theorem we have : $rk(u)+\dim \ker u = \dim \ker \mathcal G $. But $\dim \ker u = 0$, because by definition of $u$, $\ker u \subset  \ker \mathcal L$ and $\ker u \subset  \ker \mathcal G$, so $\dim \ker u = 0$.
3) Try to manipulate the two previous inequalities/inclusions.
4) Let $x \in \ker \mathcal G$, we have $\mathcal G (\mathcal L(x))=\mathcal L( \mathcal G(x))=\mathcal L(0)=0$. So $x \in \ker (\mathcal G \circ \mathcal L)$.
A: Hint
1) for $y\in L(\ker G)$, there is $x\in \ker G$ such that $y=L(x)$ and we have 
$L(y)=\cdots$
2) We have $\ker (L_{|\ker G})=\ker L\cap\ker G=\{0\}$ so the restriction $L_{|\ker G}$ is injective and then by the rank-nullity theorem......
3)By 1) and 2) $$\dim\ker G=\dim(L(\ker G))\le \ker L$$
and interchanging the role of $L$ and $G$ we get.....
4) For $x\in\ker G$ we have $G(L(x))=L(G(x))=\cdots$
A: I'm not sure what you are getting at with 1) What is $Im\{Ker G\}$? A different proof might go like this: 
Let $x\in L(Ker(G))$ we want to show $x\in Ker(L)$ i.e. that $L(x) = 0$. Well as $x\in L(Ker(G))$ then $x = L(y)$ for some $y\in Ker(G)$, so $L(x) = L(L(y)) = G(G(y))$. But, $y\in Ker(G)$ and so $G(y) = 0$ and so $G(G(y)) = 0$. Hence $L(x) = G(G(y)) = 0$ as desired.
